If tangents are drawn to the circle x^2+y^2=1 | vedclass

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(D) Let $(h, k)$ be the point of intersection of the tangents. The chord of contact of the tangents to the circle $x^2+y^2=12$ is given by $hx+ky=12$,

Vedclass · Accepted Answer

$\left(6, -\frac{18}{5}\right)$

Vedclass · Answer

(D) Let $(h, k)$ be the point of intersection of the tangents. The chord of contact of the tangents to the circle $x^2+y^2=12$ is given by $hx+ky=12$,or $hx+ky-12=0$. This chord of contact is the common chord of the two circles. The equation of the common chord is obtained by subtracting the two circle equations: $(x^2+y^2-12) - (x^2+y^2-5x+3y-2) = 0$,which simplifies to $5x-3y-10=0$. Since $hx+ky-12=0$ and $5x-3y-10=0$ represent the same line,their coefficients must be proportional: $\frac{h}{5} = \frac{k}{-3} = \frac{-12}{-10} = \frac{6}{5}$. Thus,$h = 5 	imes \frac{6}{5} = 6$ and $k = -3 	imes \frac{6}{5} = -\frac{18}{5}$. The point of intersection is $\left(6, -\frac{18}{5}ight)$.

If tangents are drawn to the circle $x^2+y^2=12$ at the points where it intersects the circle $x^2+y^2-5x+3y-2=0$,then the coordinates of the point of intersection of those tangents are

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